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May 6th, 2019, 07:26 AM   #11
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 Originally Posted by Integrator Hello, I suppose we have to consider the possibility that $\displaystyle x = a + k(a)i$ where $\displaystyle a \in \mathbb R$ and $\displaystyle i^2=-1$ , which means that the initial equation will result an equation in the form $\displaystyle u + vi=0$ where $\displaystyle u=f(a)$ and $\displaystyle v=g(a)$.How do we solve the equation $\displaystyle f(a)+g(a)i=0$ where $\displaystyle i^2=-1$?Thank you very much! All the best, Integrator
Sorry, but I have no idea how one would find complex solutions for the initial equation you provided. Hopefully, someone else with the requisite knowledge can offer you some guidance.

 May 10th, 2019, 09:11 AM #12 Member   Joined: Aug 2018 From: România Posts: 45 Thanks: 2 Hello all, Solving the WolframAlpha computing program: https://www.wolframalpha.com/input/?...%2B2%7C%27%3D0 ----------------------------------------- The first two solutions given by "WolframAlpha" are the same as those found by me. From my calculations would also result solutions for $\displaystyle k(a)=0$ , $\displaystyle k(a)=i(a-2)$ , $\displaystyle k(a)=i(a-1)$ and $\displaystyle k(a)=\frac{i(2a-3)}{2}$ where $\displaystyle i^2=-1$ , $\displaystyle a=Re(x)$ and $\displaystyle k(a)=Im(x)$ , but I do not know how to interpret them to calculate $\displaystyle x=a+k(a)i$ .... All the best, Integrator Last edited by Integrator; May 10th, 2019 at 09:34 AM.

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